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electrical_engineering_and_electronics_2:block01 [2026/03/05 01:43] – angelegt mexleadminelectrical_engineering_and_electronics_2:block01 [2026/04/06 23:36] (current) – mexleadmin
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 ===== Learning Objectives ===== ===== Learning Objectives =====
 <callout> <callout>
-After this 90-minute block, you can +After this 90-minute block, you  
-  * ...+  * know the time constant $\tau$ and in particularly calculate it. 
 +  * determine the time characteristic of the currents and voltages at the RC element for a given resistance and capacitance. 
 +  * know the continuity conditions of electrical quantities. 
 +  * know when (=according to which measure) the capacitor is considered to be fully charged/discharged, i.e. a steady state can be considered to have been reached. 
 +  * can calculate the energy content in a capacitor. 
 +  * can calculate the change in energy of a capacitor resulting from a change in voltage between the capacitor terminals. 
 +  * can calculate (initial) current, (final) voltage, and charge when balancing the charge of several capacitors (also via resistors).
 </callout> </callout>
 +
  
 ===== Preparation at Home ===== ===== Preparation at Home =====
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 </callout> </callout>
  
-===== 5.1 Time Course of the Charging and Discharging Process =====+====Time Course of the Charging and Discharging Process ====
  
-<callout> 
- 
-=== Learning Objectives === 
- 
-By the end of this section, you will be able to: 
-  - know the time constant $\tau$ and in particularly calculate it. 
-  - determine the time characteristic of the currents and voltages at the RC element for a given resistance and capacitance. 
-  - know the continuity conditions of electrical quantities. 
-  - know when (=according to which measure) the capacitor is considered to be fully charged/discharged, i.e. a steady state can be considered to have been reached. 
- 
-</callout> 
  
 In the simulation below you can see the circuit mentioned above in a slightly modified form: In the simulation below you can see the circuit mentioned above in a slightly modified form:
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 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
- 
-Here is a short introduction about the transient behavior of an RC element (starting at 15:07 until 24:55) 
-{{youtube>8nyNamrWcyE?start=907&stop=1495}} 
  
 To understand the charging process of a capacitor, an initially uncharged capacitor with capacitance $C$ is to be charged by a DC voltage source $U_{\rm s}$ via a resistor $R$. To understand the charging process of a capacitor, an initially uncharged capacitor with capacitance $C$ is to be charged by a DC voltage source $U_{\rm s}$ via a resistor $R$.
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 C = {{q(t)}  \over{u_C(t)}} = {{{\rm d}q}  \over{{\rm d}u_C}} = {\rm const.} \tag{5.1.1}  C = {{q(t)}  \over{u_C(t)}} = {{{\rm d}q}  \over{{\rm d}u_C}} = {\rm const.} \tag{5.1.1} 
 \end{align*} \end{align*}
- 
-The following explanations are also well explained in these two videos on [[https://www.youtube.com/watch?v=csFh588BODY&ab_channel=MattAnderson|charging]] and [[https://www.youtube.com/watch?v=eCOLkUPSpxc&ab_channel=lasseviren1|discharging]]. 
- 
  
 ==== Charging a capacitor at time t=0 ==== ==== Charging a capacitor at time t=0 ====
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 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-===== 5.2 Energy stored in a Capacitor ===== +==== Energy stored in a Capacitor ====
- +
-<callout> +
- +
-=== Learning Objectives === +
- +
-By the end of this section, you will be able to: +
-  - calculate the energy content in a capacitor. +
-  - calculate the change in energy of a capacitor resulting from a change in voltage between the capacitor terminals. +
-  - calculate (initial) current, (final) voltage, and charge when balancing the charge of several capacitors (also via resistors). +
- +
-</callout>+
  
 <WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2.svg}} </WRAP> <WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2.svg}} </WRAP>
  
-Now the capacitor as energy storage is to be looked at more closely. This derivation is also explained in [[https://www.youtube.com/watch?v=PTyB6_Kt_5A&ab_channel=TheOrganicChemistryTutor|this youtube video]]. For this, we consider again the circuit in <imgref imageNo02 > an. According to the chapter [[:electrical_engineering_1:preparation_properties_proportions#power_and_efficiency|Preparation, Properties, and Proportions]], the power for constant values (DC) is defined as:+Now the capacitor as energy storage is to be looked at more closely. For this, we consider again the circuit in <imgref imageNo02 > an. According to the chapter [[:electrical_engineering_1:preparation_properties_proportions#power_and_efficiency|Preparation, Properties, and Proportions]], the power for constant values (DC) is defined as:
  
 \begin{align*} \begin{align*}
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 ==== Worked examples ==== ==== Worked examples ====
  
-...+ 
 +<panel type="info" title="Exercise 1.1 Capacitor charging/discharging practice Exercise ">  
 +<WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
 + 
 +{{youtube>b_S-Ni5n-2s}} 
 + 
 +</WRAP></WRAP></panel> 
 + 
 +#@TaskTitle_HTML@# 1.2 Capacitor charging/discharging #@TaskText_HTML@# 
 + 
 +The following circuit shows a charging/discharging circuit for a capacitor. 
 + 
 +The values of the components shall be the following: 
 +  * $R_1 = 1.0 \rm k\Omega$ 
 +  * $R_2 = 2.0 \rm k\Omega$ 
 +  * $R_3 = 3.0 \rm k\Omega$ 
 +  * $C   = 1 \rm \mu F$ 
 +  * $S_1$ and $S_2$ are opened in the beginning (open-circuit) 
 + 
 +{{drawio>electrical_engineering_1:Exercise522setup.svg}} 
 + 
 +1. For the first tasks, the switch $S_1$ gets closed at $t=t_0 = 0s$. \\ 
 + 
 +1.1 What is the value of the time constant $\tau_1$? 
 + 
 +#@HiddenBegin_HTML~Solution1,Solution~@# 
 + 
 +The time constant $\tau$ is generally given as: $\tau= R\cdot C$. \\ 
 +Now, we try to determine which $R$ and $C$ must be used here. \\ 
 +To find this out, we have to look at the circuit when $S_1$ gets closed. 
 + 
 +{{drawio>electrical_engineering_1:Exercise522sol1.svg}} 
 + 
 +We see that for the time constant, we need to use $R=R_1 + R_2$. 
 + 
 +#@HiddenEnd_HTML~Solution1,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result1,Result~@# 
 +\begin{align*} 
 +\tau_1 &= R\cdot C \\ 
 +       &= (R_1 + R_2) \cdot C \\ 
 +       &= 3~\rm ms \\ 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Result1,Result~@# 
 + 
 +1.2 What is the formula for the voltage $u_{R2}$ over the resistor $R_2$? Derive a general formula without using component values! 
 + 
 +#@HiddenBegin_HTML~Solution2,Solution~@# 
 + 
 +To get a general formula, we again look at the circuit, but this time with the voltage arrows. 
 + 
 +{{drawio>electrical_engineering_1:Exercise522sol2.svg}} 
 + 
 +We see, that: $U_1 = u_C + u_{R2}$ and there is only one current in the loop: $i = i_C = i_{R2}$\\ 
 +The current is generally given with the exponential function: $i_c = {{U}\over{R}}\cdot e^{-t/\tau}$, with $R$ given here as $R = R_1 + R_2$. 
 +Therefore, $u_{R2}$ can be written as: 
 + 
 +\begin{align*} 
 +u_{R2} &= R_2 \cdot i_{R2} \\ 
 +       &= U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{-t/ \tau}  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Solution2,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result2,Result~@# 
 +\begin{align*} 
 +u_{R2} = U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{t/ \tau} 
 +\end{align*} 
 +#@HiddenEnd_HTML~Result2,Result~@# 
 + 
 +2. At a distinct time $t_1$, the voltage $u_C$ is charged up to $4/5 \cdot U_1$. 
 +At this point, the switch $S_1$ will be opened. \\ Calculate $t_1$! 
 + 
 +#@HiddenBegin_HTML~Solution3,Solution~@# 
 + 
 +We can derive $u_{C}$ based on the exponential function: $u_C(t) = U_1 \cdot (1-e^{-t/\tau})$. \\ 
 +Therefore, we get $t_1$ by: 
 + 
 +\begin{align*} 
 +u_C = 4/5 \cdot U_1              &=  U_1 \cdot (1-e^{-t/\tau}) \\ 
 +      4/5                        &=             1-e^{-t/\tau} \\ 
 +      e^{-t/\tau}                &=             1-4/5 = 1/5\\ 
 +         -t/\tau                 &=             \rm ln (1/5) \\ 
 +          t                      &= -\tau \cdot \rm ln (1/5) \\ 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Solution3,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result3,Result~@# 
 +\begin{align*} 
 +          t                      &=  3~{\rm ms} \cdot 1.61 \approx 4.8~\rm ms \\ 
 +\end{align*} 
 +#@HiddenEnd_HTML~Result3,Result~@# 
 + 
 +3. The switch $S_2$ will get closed at the moment $t_2 = 10 ~\rm ms$. The values of the voltage sources are now: $U_1 = 5.0 ~\rm V$ and $U_2 = 10 ~\rm V$. 
 + 
 +3.1 What is the new time constant $\tau_2$? 
 + 
 +#@HiddenBegin_HTML~Solution4,Solution~@# 
 + 
 +Again, the time constant $\tau$ is given as: $\tau= R\cdot C$. \\ 
 +Again, we try to determine which $R$ and $C$ must be used here. \\ 
 +To find this out, we have to look at the circuit when $S_1$ is open and $S_2$ is closed. 
 + 
 +{{drawio>electrical_engineering_1:Exercise522sol4.svg}} 
 + 
 +We see that for the time constant, we now need to use $R=R_3 + R_2$. 
 + 
 +\begin{align*} 
 +\tau_2 &= R\cdot C \\ 
 +       &= (R_3 + R_2) \cdot C \\ 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Solution4,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result4,Result~@# 
 +\begin{align*} 
 +\tau_2 &= 5~\rm ms \\ 
 +\end{align*} 
 +#@HiddenEnd_HTML~Result4,Result~@# 
 + 
 +3.2 Calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$. 
 + 
 +#@HiddenBegin_HTML~Solution5,Solution~@# 
 + 
 +To calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$, we first have to find out the value of $u_{R2}(t_2 = 10 ~\rm ms)$, when $S_2$ just got closed. \\ 
 +  * Starting from $t_2 = 10 ~\rm ms$, the voltage source $U_2$ charges up the capacitor $C$ further. 
 +  * Before at $t_1$, when $S_1$ got opened, the value of $u_c$ was: $u_c(t_1) = 4/5 \cdot U_1 = 4 ~\rm V$. 
 +  * This is also true for $t_2$, since between $t_1$ and $t_2$ the charge on $C$ does not change: $u_c(t_2) = 4 ~\rm V$. 
 +  * In the first moment after closing $S_2$ at $t_2$, the voltage drop on $R_3 + R_2$ is: $U_{R3+R2} = U_2 - u_c(t_2) = 6 ~\rm V$. 
 +  * So the voltage divider of $R_3 + R_2$ lead to $ \boldsymbol{u_{R2}(t_2 = 10 ~\rm ms)} =  {{R_2}\over{R_3 + R2}} \cdot U_{R3+R2} = {{2 {~\rm k\Omega}}\over{3 {~\rm k\Omega} + 2 {~\rm k\Omega} }} \cdot 6 ~\rm V =  \boldsymbol{2.4 ~\rm V} $ 
 + 
 +We see that the voltage on $R_2$ has to decrease from $2.4 ~\rm V $ to $1/10 \cdot U_2 = 1 ~\rm V$. \\ 
 +To calculate this, there are multiple ways. In the following, one shall be retraced: 
 +  * We know, that the current $i_C = i_{R2}$ subsides exponentially: $i_{R2}(t) = I_{R2~ 0} \cdot {\rm e}^{-t/\tau}$ 
 +  * So we can rearrange the task to focus on the change in current instead of the voltage. 
 +  * The exponential decay is true regardless of where it starts. 
 + 
 +So from ${{i_{R2}(t)}\over{I_{R2~ 0}}} =  {\rm e}^{-t/\tau}$ we get  
 +\begin{align*} 
 +{{i_{R2}(t_3)}\over{i_{R2}(t_2)}} &=                                 {\rm exp} \left( -{{t_3 - t_2}\over{\tau_2}}       \right) \\ 
 +-{{t_3 - t_2}\over{\tau_2}}       &=                                 {\rm ln } \left( {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} \right) \\ 
 +   t_3                            &= t_2          - \tau_2     \cdot {\rm ln } \left( {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} \right) \\ 
 +   t_3                            &= 10 ~{\rm ms} - 5~{\rm ms} \cdot {\rm ln } \left( {{1 ~\rm V   }\over{2.4 ~\rm V }} \right) \\ 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Solution5,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result5,Result~@# 
 +\begin{align*} 
 +t_3 &= 14.4~\rm ms \\ 
 +\end{align*} 
 +#@HiddenEnd_HTML~Result5,Result~@# 
 + 
 +3.3 Draw the course of time of the voltage $u_C(t)$ over the capacitor. 
 + 
 +{{drawio>electrical_engineering_1:Exercise522task6.svg}} 
 + 
 + 
 +#@HiddenBegin_HTML~Result6,Result~@# 
 +{{drawio>electrical_engineering_1:Exercise522sol6.svg}} 
 +#@HiddenEnd_HTML~Result6,Result~@# 
 + 
 +#@TaskEnd_HTML@# 
 + 
 +#@TaskTitle_HTML@#1.3 Charge balance of two capacitors \\ <fs medium>(educational exercise, not part of an exam)</fs>#@TaskText_HTML@# 
 + 
 + 
 +<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3EZgBwDZkCYAsBWAnDgOxYDMG52IJEOIpIOApgLRhgBQAxuBsnX2F7hCqOlFjxIeaTNly84sNMKRsYQmFWFCyNOEjjJkDtxI4M-KueGishyfMcyQeaMjgY8hHJmR5UFgYwRhwA5jwCIhHgeIFQ7OFmFuqiSTFxkOwYkITRWAJCYLYgAK5gABQAwmAAlFk5Vhb5jeDFZVUYdQDOIOR2ZL2EFgMGEABmAIYANl2M7ADueZG2fJkkKlSQds1pzYjxAMo2ln2WQSCTM4xQvewATr2UqHYYQyDPNxyLrxYfJFtnBaDYbDAHNTLQEAAUQAdow7qEAJ4AW0YABd4TcLMMQJVeliQABVcAkiwAdTEBhIIAAapUaQcoVNkQAdLpgNkYNkkNlYNk4Fl3ADidwA9iUYQATJmsroC4ViiXS5ncqAs7nqrowHAAKnKE2YnBqAD0MHqAEaGmpsyCa21au3sLCaBiCKgWHCoAwfCCw+FI1EYh72nla6C6-VW00Wq3gG1xrXsABKrpB-CpGToQQk8CcThu2Zw9VyODdYCw23c4ArYnalU6xdT70r3u2pQq1TqKdLyTd+VGbvOWGzRjzjgL4iLzoMPfAbs9oxrvrhCJR6MxIZt4b1BqN0fKlqNCdGNvYAA8qP99NSSMP9MUULcL9RqXg+CQr5oIC87JVzyBkFyL99DsTQ20oP8LzA+851AqQxEfDgL2wAw8BeIgXFyH8QCTf8MBwalaAwVBqT4UC+AwPCSFEAx8JxbCDjw1ReiIt0hHI8B-xQIjaOMXpqQY-9cD4ZhXneUsQFEtsihAThpk4EopjktEAEtRRhNlRTGNk0QAC0YNlGBXUIVMYLp2EhP1Vy6EpkVRYN8RxMkIH2clsVJEBAEQiCk7CpWl6UZFV2U5DUuj5OVBRFcUpRlDUFWi5VZVoW0NTZSMjXVdUAAcDytG07Xy09pybJQPRrbIBGhYzERsuyN1C9LrRZHLDxqY9j0bL131+O8BjseYO1qTrIAEN1ngHJoQHmDAOjqbJci69Jm1GWIxE8+ZG1nUr0yW-rBrmhpZ1OftHjbabZvYc06GaZhmlQVBcnw3JMlFBCbjPXgv2SbNXIA3pwxufpDHw+JXtofYPuQL79EUDyKIBmcqGB5L2Fez93s+4xAgkRBoAgdzVhzEGqT84JidRqgXIxqGsfsXGIGpCxCfgYnKaRsmUbR5Jqeh4I+KQMQmfsVRkvddnYHJkhVhAAAxfHSQMVgQAASRhFTVOmNkaVFKY0QmUI5iAA noborder}} </WRAP> 
 + 
 +In the simulation, you see the two capacitors $C_1$ and $C_2$ (The two small resistors with $1 ~\rm µ\Omega$ have to be there for the simulation to run). At the beginning, $C_1$ is charged to $10~{\rm V}$ and $C_2$ to $0~{\rm V}$. With the switches $S_1$ and $S_2$ you can choose whether 
 + 
 +  - the capacitances $C_1$ and $C_2$ are shorted, or 
 +  - the capacitors $C_1$ and $C_2$ are connected via resistor $R$. 
 + 
 +On the right side of the simulation, there are some additional "measuring devices" to calculate the stored potential energy from the voltages across the capacitors. 
 + 
 +In the following, the charging and discharging of a capacitor are to be explained with this construction. 
 + 
 +~~PAGEBREAK~~ ~~CLEARFIX~~  
 + 
 +Under the electrical structure, the following quantities are shown over time: 
 + 
 +^Voltage $u_1(C_1)$ of the first capacitor^Voltage $u_2(C_2)$ of the second capacitor^Stored energy $w_1(C_1)$^Stored energy $w_2(C_2)$^Total energy $\sum w$| 
 +|Initially charged to $10~{\rm V}$|Initially neutrally charged ($0~{\rm V}$)|Initially holds: \\ $w_1(C_1)= {1 \over 2} \cdot C \cdot U^2 = {1 \over 2} \cdot 10~{\rm µF} \cdot (10~{\rm V})^2 = 500~{\rm µW}$ \\ In the oscilloscope, equals $1~{\rm V} \sim 1~{\rm W}$|Initially, $w_2(C_2)=0$ , since the capacitor is not charged.|The total energy is $w_1 + w_2 = w_1$| 
 + 
 +The capacitor $C_1$ has thus initially stored the full energy and via closing of the switch, $S_2$ one would expect a balancing of the voltages and an equal distribution of the energy $w_1 + w_2 = 500~\rm µW$. 
 + 
 +  - Close the switch $S_2$ (the toggle switch $S_1$ should point to the switch $S_2$). What do you find? 
 +      - What do the voltages $u_1$ and $u_2$ do? 
 +      - What are the energies and the total energy? \\ How is this understandable with the previous total energy? 
 +  - Open $S_2$ - the changeover switch $S_1$ should not be changed. What do you find? 
 +      - What do the voltages $u_1$ and $u_2$ do? 
 +      - What are the energies and the total energy? \\ How is this understandable with the previous total energy? 
 +  - Repeat 1. and 2. several times. Can anything be deduced regarding the distribution of energy? 
 +  - Change the switch $S_2$ to the resistor. What do you observe? 
 +      - What do the voltages $u_1$ and $u_2$ do? 
 +      - What are the energies and the total energy? \\ How is this understandable with the previous total energy? 
 + 
 +#@TaskEnd_HTML@# 
 + 
 + 
 +#@TaskTitle_HTML@#1.4  Machine-Vision Strobe Unit: Charging and Safe Discharge of a Flash Capacitor                    #@TaskText_HTML@# 
 + 
 +A machine-vision inspection system on a production line uses a short high-voltage flash pulse. 
 +For this purpose, an energy-storage capacitor is charged from a DC source and must be safely discharged before maintenance. 
 + 
 +Data: 
 +\begin{align*} 
 +C &= 1~{\rm \mu F} \\ 
 +W_e &= 0.1~{\rm J} \\ 
 +I_{\rm max} &= 100~{\rm mA} \\ 
 +R_i &= 10~{\rm M\Omega} 
 +\end{align*} 
 + 
 +1. What voltage must the capacitor have so that it stores the required energy? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~101~@# 
 +<WRAP leftalign> 
 +\begin{align*} 
 +W_e &= \frac{1}{2} C U^2 \\ 
 +U &= \sqrt{\frac{2W_e}{C}} 
 +   = \sqrt{\frac{2 \cdot 0.1~{\rm J}}{1 \cdot 10^{-6}~{\rm F}}} \\ 
 +  &= \sqrt{200000}~{\rm V} 
 +   \approx 447.2~{\rm V} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~101~@# 
 +\begin{align*} 
 +U = 447.2~{\rm V} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +2. The charging current must not exceed $100~\rm mA$ at the start of charging. What charging resistor is required? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~102~@# 
 +<WRAP leftalign> 
 +At the beginning of charging, the capacitor behaves like a short circuit, so 
 +\begin{align*} 
 +i_{C{\rm max}} = i_C(t=0) = \frac{U}{R} 
 +\end{align*} 
 +Thus, 
 +\begin{align*} 
 +R &\ge \frac{U}{I_{\rm max}} 
 +   = \frac{447.2~{\rm V}}{0.1~{\rm A}} \\ 
 +  &\approx 4472~{\rm \Omega} 
 +   = 4.47~{\rm k\Omega} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~102~@# 
 +\begin{align*} 
 +R \ge 4.47~{\rm k\Omega} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +3. How long does the charging process take until the capacitor is practically fully charged? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~103~@# 
 +<WRAP leftalign> 
 +The time constant is 
 +\begin{align*} 
 +T = RC = 4.47~{\rm k\Omega} \cdot 1~{\rm \mu F} = 4.47~{\rm ms} 
 +\end{align*} 
 +In engineering practice, a capacitor is considered practically fully charged after about $5T$: 
 +\begin{align*} 
 +t \approx 5T = 5 \cdot 4.47~{\rm ms} = 22.35~{\rm ms} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~103~@# 
 +\begin{align*} 
 +t \approx 22.35~{\rm ms} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +4. Give the time-dependent capacitor voltage and the voltage across the charging resistor. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~104~@# 
 +<WRAP leftalign> 
 +For the charging process: 
 +\begin{align*} 
 +u_C(t) &= U\left(1-e^{-t/T}\right) \\ 
 +u_R(t) &= Ue^{-t/T} 
 +\end{align*} 
 +with 
 +\begin{align*} 
 +U &= 447.2~{\rm V} \\ 
 +T &= 4.47~{\rm ms} 
 +\end{align*} 
 +So the capacitor voltage rises exponentially from $0$ to $447.2~\rm V$, while the resistor voltage falls exponentially from $447.2~\rm V$ to $0$. 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~104~@# 
 +\begin{align*} 
 +u_C(t) &= 447.2\left(1-e^{-t/4.47{\rm ms}}\right)~{\rm V} \\ 
 +u_R(t) &= 447.2\,e^{-t/4.47{\rm ms}}~{\rm V} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +5. After charging, the capacitor is disconnected from the source. Its leakage can be modeled by an internal resistance of $10~\rm M\Omega$. 
 +After what time has the stored energy dropped to one half, and what is the capacitor voltage then? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~105~@# 
 +<WRAP leftalign> 
 +Half the energy means 
 +\begin{align*} 
 +W_e' = 0.5W_e 
 +\end{align*} 
 +Since 
 +\begin{align*} 
 +W_e = \frac{1}{2}CU^2 
 +\end{align*} 
 +the voltage at half energy is 
 +\begin{align*} 
 +U' = \frac{U}{\sqrt{2}} 
 +    = \frac{447.2~{\rm V}}{\sqrt{2}} 
 +    = 316.2~{\rm V} 
 +\end{align*} 
 +For the discharge through the internal resistance: 
 +\begin{align*} 
 +u_C(t) = Ue^{-t/T_2} 
 +\end{align*} 
 +with 
 +\begin{align*} 
 +T_2 = R_iC = 10~{\rm M\Omega} \cdot 1~{\rm \mu F} = 10~{\rm s} 
 +\end{align*} 
 +Set $u_C(t)=U'$: 
 +\begin{align*} 
 +Ue^{-t/T_2} &= U' \\ 
 +t &= T_2 \ln\left(\frac{U}{U'}\right) \\ 
 +  &= 10~{\rm s}\cdot\ln\left(\frac{447.2}{316.2}\right) \\ 
 +  &\approx 3.47~{\rm s} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~105~@# 
 +\begin{align*} 
 +U' = 316.2~{\rm V} \\ 
 +t = 3.47~{\rm s} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +6. The fully charged capacitor is discharged through the charging resistor before maintenance. 
 +How long does the discharge take, and how much energy is converted into heat in the resistor? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~106~@# 
 +<WRAP leftalign> 
 +The discharge time constant through the same resistor is again 
 +\begin{align*} 
 +T = RC = 4.47~{\rm ms} 
 +\end{align*} 
 +Thus the practical discharge time is 
 +\begin{align*} 
 +t \approx 5T = 22.35~{\rm ms} 
 +\end{align*} 
 +The complete stored capacitor energy is converted into heat in the resistor: 
 +\begin{align*} 
 +W_R = W_e = 0.1~{\rm Ws} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~106~@# 
 +\begin{align*} 
 +t \approx 22.35~{\rm ms} \\ 
 +W_R = 0.1~{\rm Ws} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +#@TaskEnd_HTML@# 
 + 
 + 
 +#@TaskTitle_HTML@#1.5 Sensor Input Buffer: Source, T-Network and Capacitor                    #@TaskText_HTML@# 
 + 
 +<wrap right>{{drawio>electrical_engineering_and_electronics_2:ExTNetworkWithCapV01.svg}}</wrap> 
 +A 12 V industrial sensor electronics unit feeds a buffered measurement node through a resistor T-network. 
 +A capacitor smooths the node voltage. At first, the load is disconnected. 
 +After the capacitor is fully charged, a measurement load is connected by a switch. 
 + 
 +Data: 
 +\begin{align*} 
 +U   &= 12~{\rm V} \\ 
 +R_1 &= 2~{\rm k\Omega} \\ 
 +R_2 &= 10~{\rm k\Omega} \\ 
 +R_3 &= 3.33~{\rm k\Omega} \\ 
 +C   &= 2~{\rm \mu F} \\ 
 +R_L &= 5~{\rm k\Omega} 
 +\end{align*} 
 + 
 +Initially, the capacitor is uncharged and the switch is open. 
 + 
 +1. What is the capacitor voltage after it is fully charged? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~201~@# 
 +<WRAP leftalign> 
 +Using the equivalent voltage source of the network on the left-hand side, the open-circuit voltage is 
 +\begin{align*} 
 +U_{0e} &= \frac{R_2}{R_1+R_2}\,U \\ 
 +       &= \frac{10~{\rm k\Omega}}{2~{\rm k\Omega}+10~{\rm k\Omega}}\cdot 12~{\rm V} \\ 
 +       &= 10~{\rm V} 
 +\end{align*} 
 +After full charging, the capacitor voltage equals this voltage. 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~201~@# 
 +\begin{align*} 
 +U_C = U_{0e} = 10~{\rm V} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +2. How long does the charging process take? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~202~@# 
 +<WRAP leftalign> 
 +The internal resistance seen by the capacitor is 
 +\begin{align*} 
 +R_{ie} &= R_3 + (R_1 \parallel R_2) \\ 
 +       &= 3.33~{\rm k\Omega} + \frac{2~{\rm k\Omega}\cdot 10~{\rm k\Omega}}{2~{\rm k\Omega}+10~{\rm k\Omega}} \\ 
 +       &= 3.33~{\rm k\Omega} + 1.67~{\rm k\Omega} \\ 
 +       &= 5.00~{\rm k\Omega} 
 +\end{align*} 
 +So the time constant is 
 +\begin{align*} 
 +T &= R_{ie}C 
 +   = 5.00~{\rm k\Omega}\cdot 2~{\rm \mu F} 
 +   = 10~{\rm ms} 
 +\end{align*} 
 +Practical charging time: 
 +\begin{align*} 
 +t \approx 5T = 50~{\rm ms} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~202~@# 
 +\begin{align*} 
 +R_{ie} = 5.00~{\rm k\Omega} \\ 
 +t \approx 50~{\rm ms} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +3. Give the time-dependent capacitor voltage. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~203~@# 
 +<WRAP leftalign> 
 +The charging law is 
 +\begin{align*} 
 +u_C(t) &= U_{0e}\left(1-e^{-t/T}\right) \\ 
 +       &= 10\left(1-e^{-t/10{\rm ms}}\right)~{\rm V} 
 +\end{align*} 
 +So the capacitor voltage rises exponentially from $0~\rm V$ to $10~\rm V$. 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~203~@# 
 +\begin{align*} 
 +u_C(t) = 10\left(1-e^{-t/10{\rm ms}}\right)~{\rm V} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +4. After the capacitor is fully charged, the switch is closed and the load resistor is connected. 
 +What is the stationary load voltage? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~204~@# 
 +<WRAP leftalign> 
 +Now use a second equivalent voltage-source step. 
 +The Thevenin source seen by the load has 
 +\begin{align*} 
 +U_{0e} &= 10~{\rm V} \\ 
 +R_{ie} &= 5.00~{\rm k\Omega} 
 +\end{align*} 
 +Thus, the stationary load voltage is 
 +\begin{align*} 
 +U_C' = U_{0e}' &= \frac{R_L}{R_{ie}+R_L}\,U_{0e} \\ 
 +               &= \frac{5~{\rm k\Omega}}{5~{\rm k\Omega}+5~{\rm k\Omega}}\cdot 10~{\rm V} \\ 
 +               &= 5~{\rm V} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~204~@# 
 +\begin{align*} 
 +U_L = 5~{\rm V} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +5. How long does it take until this new stationary state is practically reached? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~205~@# 
 +<WRAP leftalign> 
 +The new internal resistance is 
 +\begin{align*} 
 +R_{ie}' &= R_{ie}\parallel R_L \\ 
 +        &= 5.00~{\rm k\Omega}\parallel 5.00~{\rm k\Omega} \\ 
 +        &= 2.50~{\rm k\Omega} 
 +\end{align*} 
 +Hence the new time constant is 
 +\begin{align*} 
 +T' &= R_{ie}'C 
 +    = 2.50~{\rm k\Omega}\cdot 2~{\rm \mu F} 
 +    = 5~{\rm ms} 
 +\end{align*} 
 +Practical settling time: 
 +\begin{align*} 
 +t \approx 5T' = 25~{\rm ms} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~205~@# 
 +\begin{align*} 
 +R_{ie}' = 2.50~{\rm k\Omega} \\ 
 +t \approx 25~{\rm ms} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +6. Give the time-dependent load voltage after the switch is closed. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~206~@# 
 +<WRAP leftalign> 
 +At the switching instant, the capacitor voltage cannot jump. 
 +Therefore: 
 +\begin{align*} 
 +u_L(0^+) &= 10~{\rm V} \\ 
 +u_L(\infty) &= 5~{\rm V} 
 +\end{align*} 
 +The voltage therefore decays exponentially toward the new final value: 
 +\begin{align*} 
 +u_L(t) &= u_L(\infty) + \left(u_L(0^+)-u_L(\infty)\right)e^{-t/T'} \\ 
 +       &= 5 + 5e^{-t/5{\rm ms}}~{\rm V} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~206~@# 
 +\begin{align*} 
 +u_L(t) = 5 + 5e^{-t/5{\rm ms}}~{\rm V} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +#@TaskEnd_HTML@# 
 + 
 + 
 +{{tag>inductors air_core_coil magnetic_field hall_sensor transient_response current_density chapter1_1}} 
 +{{include_n>300}} 
 + 
 +#@TaskTitle_HTML@#1.6 Hall-Sensor Calibration Coil: Short Air-Core Coil                    #@TaskText_HTML@# 
 + 
 +A Hall-sensor calibration bench uses a short air-core coil to create a defined magnetic field. 
 +An air-core coil is chosen because it avoids hysteresis and remanence effects. 
 +The coil is wound as a short cylindrical coil. 
 + 
 +Data: 
 +\begin{align*} 
 +l &= 22~{\rm mm} \\ 
 +d &= 20~{\rm mm} \\ 
 +d_{\rm Cu} &= 0.8~{\rm mm} \\ 
 +N &= 25 \\ 
 +\rho_{\rm Cu,20^\circ C} &= 0.0178~{\rm \Omega\,mm^2/m} 
 +\end{align*} 
 + 
 +A DC current of $1~\rm A$ shall flow through the coil. 
 + 
 +1. Calculate the coil resistance $R$ at room temperature. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~301~@# 
 +<WRAP leftalign> 
 +The wire cross section is 
 +\begin{align*} 
 +A_{\rm Cu} &= \pi\left(\frac{d_{\rm Cu}}{2}\right)^2 
 +           = \pi(0.4~{\rm mm})^2 \\ 
 +           &= 0.503~{\rm mm^2} 
 +\end{align*} 
 +The total wire length is approximated by the number of turns times the circumference: 
 +\begin{align*} 
 +l_{\rm Cu} &= N\pi d \\ 
 +           &= 25\pi \cdot 20~{\rm mm} \\ 
 +           &= 1570.8~{\rm mm} 
 +           = 1.571~{\rm m} 
 +\end{align*} 
 +Thus, 
 +\begin{align*} 
 +R &= \rho_{\rm Cu}\frac{l_{\rm Cu}}{A_{\rm Cu}} \\ 
 +  &= 0.0178~{\rm \Omega\,mm^2/m}\cdot\frac{1.571~{\rm m}}{0.503~{\rm mm^2}} \\ 
 +  &\approx 0.0556~{\rm \Omega} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~301~@# 
 +\begin{align*} 
 +R = 55.6~{\rm m\Omega} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +2. Calculate the coil inductance $L$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~302~@# 
 +<WRAP leftalign> 
 +For this short air-core coil, use 
 +\begin{align*} 
 +L = N^2 \cdot \frac{\mu_0 A}{l}\cdot\frac{1}{1+\frac{d}{2l}} 
 +\end{align*} 
 +with 
 +\begin{align*} 
 +A &= \pi\left(\frac{d}{2}\right)^2 
 +   = \pi(10~{\rm mm})^2 
 +   = 314.16~{\rm mm^2} 
 +   = 3.1416\cdot 10^{-4}~{\rm m^2} \\ 
 +\mu_0 &= 4\pi\cdot 10^{-7}~{\rm Vs/(Am)} 
 +\end{align*} 
 +Therefore, 
 +\begin{align*} 
 +L &= 25^2 \cdot \frac{4\pi\cdot 10^{-7}\,{\rm Vs/(Am)} \cdot 3.1416\cdot 10^{-4}~{\rm m^2}}{22\cdot 10^{-3}~{\rm m}} 
 +   \cdot \frac{1}{1+\frac{20}{2\cdot 22}} \\ 
 +  &\approx 7.71\cdot 10^{-6}~{\rm H} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~302~@# 
 +\begin{align*} 
 +L = 7.71~{\rm \mu H} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +3. Which DC voltage must be applied so that the stationary current becomes $I=1~\rm A$? 
 +How large is the current density $j$ in the copper wire? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~303~@# 
 +<WRAP leftalign> 
 +In the stationary DC state, the coil behaves like its ohmic resistance: 
 +\begin{align*} 
 +U &= RI \\ 
 +  &= 55.6~{\rm m\Omega}\cdot 1~{\rm A} \\ 
 +  &= 55.6~{\rm mV} 
 +\end{align*} 
 +The current density is 
 +\begin{align*} 
 +j &= \frac{I}{A_{\rm Cu}} \\ 
 +  &= \frac{1~{\rm A}}{0.503~{\rm mm^2}} \\ 
 +  &\approx 1.99~{\rm A/mm^2} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~303~@# 
 +\begin{align*} 
 +U = 55.6~{\rm mV} \\ 
 +j = 1.99~{\rm A/mm^2} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +4. How much magnetic energy is stored in the coil in the stationary state? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~304~@# 
 +<WRAP leftalign> 
 +\begin{align*} 
 +W_m &= \frac{1}{2}LI^2 \\ 
 +    &= \frac{1}{2}\cdot 7.71\cdot 10^{-6}~{\rm H}\cdot (1~{\rm A})^2 \\ 
 +    &= 3.86\cdot 10^{-6}~{\rm Ws} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~304~@# 
 +\begin{align*} 
 +W_m = 3.86\cdot 10^{-6}~{\rm Ws} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +5. Give the time-dependent coil current $i(t)$ when the coil is switched on. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~305~@# 
 +<WRAP leftalign> 
 +A coil current cannot jump instantly. 
 +It starts at $0$ and approaches the final value $I=1~\rm A$ exponentially: 
 +\begin{align*} 
 +i(t) = I\left(1-e^{-t/T}\right) 
 +\end{align*} 
 +So the sketch starts at $0~\rm A$, rises quickly, and then slowly approaches $1~\rm A$. 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~305~@# 
 +\begin{align*} 
 +i(t) = 1\left(1-e^{-t/T}\right)~{\rm A} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +6. How long does it take until the current has practically reached its stationary value? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~306~@# 
 +<WRAP leftalign> 
 +The time constant is 
 +\begin{align*} 
 +T &= \frac{L}{R} 
 +   = \frac{7.71~{\rm \mu H}}{55.6~{\rm m\Omega}} \\ 
 +  &\approx 138.9~{\rm \mu s} 
 +\end{align*} 
 +A practical final value is reached after about $5T$: 
 +\begin{align*} 
 +t \approx 5T = 5\cdot 138.9~{\rm \mu s} \approx 695~{\rm \mu s} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~306~@# 
 +\begin{align*} 
 +t \approx 695~{\rm \mu s} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +7. How much energy is dissipated as heat in the coil resistance during the current build-up? 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~307~@# 
 +<WRAP leftalign> 
 +Using the current from task 5, 
 +\begin{align*} 
 +i(t) = I\left(1-e^{-t/T}\right) 
 +\end{align*} 
 +the heat dissipated in the winding resistance up to the practical final time $5T$ is 
 +\begin{align*} 
 +W_R &= \int_0^{5T} R\,i^2(t)\,dt \\ 
 +    &= R I^2 \int_0^{5T} \left(1-e^{-t/T}\right)^2 dt 
 +\end{align*} 
 +For this interval, the integral is approximately 
 +\begin{align*} 
 +\int_0^{5T} \left(1-e^{-t/T}\right)^2 dt \approx \frac{7}{2}T 
 +\end{align*} 
 +Thus, 
 +\begin{align*} 
 +W_R &\approx RI^2\cdot \frac{7}{2}T \\ 
 +    &= 0.0556~{\rm \Omega}\cdot (1~{\rm A})^2 \cdot \frac{7}{2}\cdot 138.9~{\rm \mu s} \\ 
 +    &\approx 27.05\cdot 10^{-6}~{\rm Ws} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~307~@# 
 +\begin{align*} 
 +W_R \approx 27.05\cdot 10^{-6}~{\rm Ws} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +#@TaskEnd_HTML@# 
 + 
 + 
 +{{page>aufgabe_7.2.6_mit_rechnung&nofooter}} 
  
 ===== Embedded resources ===== ===== Embedded resources =====
 +<WRAP>
 <WRAP column half> <WRAP column half>
-Explanation (video): ...+Here is a short introduction about the transient behavior of an RC element (starting at 15:07 until 24:55) 
 +{{youtube>8nyNamrWcyE?start=907&stop=1495}}
 </WRAP> </WRAP>
  
-~~PAGEBREAK~~ ~~CLEARFIX~~+<WRAP column half> 
 +Mathematical explanation of charging a capacitor 
 +{{youtube>csFh588BODY}} 
 +</WRAP> 
 +</WRAP> 
 +\\ \\ \\ \\ 
 +<WRAP> 
 +<WRAP column half> 
 +Mathematical explanation of discharging a capacitor 
 +{{youtube>eCOLkUPSpxc}} 
 +</WRAP>
  
 +<WRAP column half>
 +Mathematical explanation of the energy stored in the capacitor
 +{{youtube>PTyB6_Kt_5A}}
 +</WRAP>
 +</WRAP>
  
 +~~PAGEBREAK~~ ~~CLEARFIX~~